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It is true that if $f$ is a real valued function, and if the iterated limits $\lim_{x\rightarrow x_0}[\lim_{y\rightarrow y_0}f(x,y)]$ and $\lim_{y\rightarrow y_0}[\lim_{x\rightarrow x_0}f(x,y)]$ exist and are equal, we cannot say anything about the double limit $\lim_{(x,y)\rightarrow (x_0,y_0)} f(x,y)$ .

But look at this proof. Let $\lim_{y\rightarrow y_0}f(x,y)=l_{y_0}(x)$, $\lim_{x\rightarrow x_0}f(x,y) = l_{x_0}(y)$, $\lim_{y\rightarrow y_0}[\lim_{x\rightarrow x_0}f(x,y)] = l_{x_0 y_0}$

Let $\epsilon$> 0. $\exists \ \delta _1 >0$ st whenever $|x-x_0|< \delta_1$ we have $|f(x,y)-l_{x_0}(y)|<\epsilon$ and $\exists$ $\delta _2 >0$ st whenever $|y-y_0|< \delta_2$ we have $|l_{x_0 y_0}-l_{x_0}(y)|<\epsilon$.

Then we have $|f(x,y)-l_{x_0 y_0}|\leq |f(x,y)-l_{x_0}(y)|+ |l_{x_0 y_0}-l_{x_0}(y)|<2\epsilon$ for $|x-x_0|<\delta_3$ and $|y-y_0|<\delta_3$ where $\delta_3= min(\delta_1, \delta_2)$

We can proceed similarly with $\lim_{x\rightarrow x_0}[\lim_{y\rightarrow y_0}f(x,y)]$ and get the same result (although it is not necessary). Thus ,we have shown the double limit is indeed the same as the iterated limit. Where did I go wrong in the proof?

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    $\begingroup$ Your definition of $l_{y_0}$ is unclear. Where did the $x$ go? The expression $\lim_{y\rightarrow y_0} f(x,y)$ assumes the limit exists for all $x$, and in general the limit would depend on $x$. So it would make more sense to say $\lim_{y\rightarrow y_0} f(x,y) = l_{y_0, x}$ (assuming the limit exists). Example: If $f(x,y)=xy$ then $\lim_{y\rightarrow y_0} f(x,y)= xy_0$. $\endgroup$
    – Michael
    Oct 27, 2020 at 13:26
  • $\begingroup$ @Michael I have made the necessary edits. Thanks for pointing that out. $\endgroup$
    – suraj
    Oct 27, 2020 at 13:41
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    $\begingroup$ As mentioned in an answer below, you really have $\delta_1(y)$ and $\delta_2(x)$. You would need to again rewrite the proof. Also, in your paragraph "Then we have ...", it is not clear what assumptions you make on $x$ and $y$. $\endgroup$
    – Michael
    Oct 27, 2020 at 14:13
  • $\begingroup$ @Michael I think I get it now. Thanks. $\endgroup$
    – suraj
    Oct 27, 2020 at 14:19

1 Answer 1

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In your proof, $\delta_1$ depends on $y$. So, it may happen that $\delta_1$ shrinks to $0$, as $y\to y_0$. Consider the function $f\colon \mathbb{R}^2\to \mathbb{R}$,

$f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ for $x,y\neq 0$ and $f(0,0):=0$.

Let $(x_0,y_0):=(0,0)$. Now, it is easy to see, that (in your notation) $l_{x_0}=l_{y_0}=l_{x_0 y_0}=0$, but $f(\delta,\delta)=1$ for any $\delta>0$, so that $\lim_{(x,y)\to(x_0,y_0)}f(x,y)$ does not exist (note that $f(0,0)=0$ by definition). Now, why does your proof don't work in this example? Well, for $0<\varepsilon<1$, no matter how small you choose $\delta_1>0$, we have for $x$ such that $|x-x_0|<\delta_1$ and $y=x$

$|f(x,y)-l_{x_0}(y)|=|f(x,y)|=1>\varepsilon$.

So in this example, if $y$ is of the same magnitude as $x$, then you cannot ensure $|f(x,y)-l_{x_0}(y)|<\varepsilon$ for $|x-x_0|<\delta_1$. You can only ensure it if you fix $y$, i.e., $\delta_1=\delta_1(y)$ depends on $y$ and $\delta_1(y)\to 0$ as $y\to y_0$ so that $\delta_3(y)=\min\{\delta_1(y), \delta_2\}\to 0$ as $y\to y_0$, contradicting the requirement that $\delta_3$ has to be strictly positive.

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  • $\begingroup$ But when y is in a neighborhood of y_0, \delta_1 must be non zero $\endgroup$
    – suraj
    Oct 27, 2020 at 13:58
  • $\begingroup$ @suraj : You are stating that $\delta_1(y)>0$ for all $y$. Fine. But so what? $\endgroup$
    – Michael
    Oct 27, 2020 at 14:14
  • $\begingroup$ @suraj: I editted my answer and gave an example. I hope that clarifies more why your proof does not work. $\endgroup$
    – ym94
    Oct 27, 2020 at 14:35

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